hdu 2680 Choose the best route （dijkstra算法）

/************************************************************************/
/*     
        hdu  Arbitrage
        dijkstra算法
        题目大意：dijkstra算法，求点与点之间最短距离。因为此题的起始点不定，所以可用
        反向图来求得，终点确定，从终点出发，dijkstra算法，求出其他点到终点最小距离。
        本题数据量较大，用floyd算法超时。
*/
/************************************************************************/

#include <cstdio>
#include <cstring>
#include <string>
#include <map>
#include <algorithm>

using namespace std;

#define MIN(a,b) a<b?a:b
#define MAX 0xfffffff

const int N = 1001;
int maps[N][N];
int dj[N],vis[N];
int m,n,s,w,num,min_num;

void build_map()
{
    int p,q,t;
    for (int i = 1; i <= n; i++)
    for (int j = i; j <= n; j++)
    maps[i][j] = maps[j][i] = ((i==j)?0:MAX);

    for (int i = 1; i <= m; i++)
    {
        scanf("%d%d%d",&p,&q,&t);
        if(maps[q][p] > t) maps[q][p] = t;//这里注意构建反向图
    }

    for (int i = 1; i <= n; i++)
    dj[i] = MAX;
}

/*
//此floyd算法超时
void floyud()
{
    for (int k = 1; k <= n; k++)
    {
        for (int i = 1; i <= n; i++)
        for (int j = 1; j <= n; j++)
        maps[i][j] = MIN(maps[i][j],maps[i][k] + maps[k][j]);
    }
}
*/
void DJ()
{
    int cur = s;
    int next,min;
    dj[cur] = 0;
    while(1)
    {
        min = MAX;
        vis[cur] = 1;
        for (int i = 1; i <= n; i++)
        {
            if (vis[i] == 1)continue;
            if (dj[i] > dj[cur] + maps[cur][i])
                dj[i] = dj[cur] + maps[cur][i];
            if (dj[i] < min)
            {
                min = dj[i];
                next = i;
            }
        }
        if ( min == MAX)break;
        cur = next;
    }
}

int main()
{
    while(scanf("%d%d%d",&n,&m,&s)!= EOF )
    {
        build_map();
        memset(vis,0,sizeof(vis));
        DJ();
        min_num = MAX;
        scanf("%d",&w);
        while(w--)
        {
            scanf("%d",&num);
            if ( dj[num] < min_num)
            min_num = dj[num];
        }
        if (min_num == MAX)
        printf("-1\n");
        else printf("%d\n",min_num);
    }
    return 0;
}